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3 years ago

Which equation represents this statement nine less than three times a number is 18

Mathematics

1 answer:

vitfil [10]3 years ago

5 0

(9x3)-9=18

Thisjj bf is the answer enjoy

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Find the complete quotient when P(x) is divided by the binomial following it.

Nutka1998 [239]

Answer:

$Q(x) = x -11$

Explanation:

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3 years ago

How to simplify (2x^4)^5

Yuliya22 [10]

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Step-by-step explanation:

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3 years ago

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Percy solved the equation x2 7x 12 = 12. his work is shown below. is percy correct? explain. 1. (x 3)(x 4) = 12 2. x 3 = 12 or x

stepladder [879]

Percy solved the equation by expanding the expression on the left-hand side, and this is incorrect

<h3>How to determine the true solution?</h3>

The equation is given as:

x^2 + 7x + 12 = 12

The first step in solving the equation is to equate the equation to 0.

i.e.

x^2 + 7x + 12 - 12 = 0

This given

x^2 + 7x = 0

From the question, Percy solved the equation by expanding the expression on the left-hand side

Hence, Percy's solution is incorrect

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2 years ago

Plss This is Due today!!!!!!!!!!!!!!!

Nina [5.8K]

Answer:

maybe this will help

Step-by-step explanation:

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2 years ago

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

3 years ago